Question

A bag contains 3 white and 2 black balls and another bag contains 2 white and 4 black balls. One bag is chosen at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is white.

Answer 1

A white ball can be drawn in two mutually exclusive ways:
(I) Selecting bag I and then drawing a white ball from it
(II) Selecting bag II and then drawing a white ball from it
Let E₁, E₂ and A be the events as defined below:
E₁ = Selecting bag I
E₂ = Selecting bag II
A = Drawing a white ball
It is given that one of the bags is selected randomly.

\[\therefore P\left( E_1 \right) = \frac{1}{2} \]

\[ P\left( E_2 \right) = \frac{1}{2}\]

\[ \text{ Now } , \]

\[P\left( A/ E_1 \right) = \frac{3}{5}\]

\[P\left( A/ E_2 \right) = \frac{2}{6} = \frac{1}{3}\]

\[\text{ Using the law of total probability, we get } \]

\[\text{ Required probability}  = P\left( A \right) = P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)\]

\[ = \frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{1}{3}\]

\[ = \frac{3}{10} + \frac{1}{6}\]

\[ = \frac{9 + 5}{30} = \frac{14}{30} = \frac{7}{15}\]